I am really baffled by a seemingly easy question on differential 1-forms.
Let $U \subset \mathbb{R}^n$ and $w$ be a differential 1-form on $U$. Suppose $\int_{\phi} w = \int_{\psi} w$ whenever $\phi$ and $\psi$ begin and end at the same points. Show that $w$ is exact.
I don't even know where to start! Any help or hint would be greatly appreciated.
On
By Poincaré-Lemma it is enough to show that $w$ is closed, i.e. $d w_x=0$ for all $x\in \mathbb{R}^n$. Consider without loss of generality $x=0$ and for $E$ a two dimensional subspace of $\mathbb{R}^n$ let $\iota_E\colon E\hookrightarrow \mathbb{R}^n$ be the inclusion. Prove that if $d(\iota_E^*w)_0=0$ for all such $E$, then necessarily $dw_0=0$. Thus it suffices to consider the case $n=2$. Take balls $B_\epsilon$ around $0\in\mathbb{R}^2$ with radius $\epsilon$, then by Stokes' Theorem $$ \frac{1}{\pi \epsilon^2}\int_{\partial B_\epsilon} w=\frac{1}{\pi \epsilon^2}\int_{B_\epsilon} dw. $$ The left integral is $0$ because $\partial B_\epsilon $ is the union of two curves with same start and end point and the right hand side tends to $c$, where $dw_0=c dx\wedge dy$ as $\epsilon\rightarrow0$ (assuming some regularity for $w$, smoothness is more than enough), thus $dw_0=0$.
Here is a good place to start. Choose some $x_0\in U$. Then, given $x\in U$, choose a path $\gamma$ with $$\gamma(0)=x_0,\quad\gamma(1)=x,$$ and set $$F(x):=\int_\gamma w.$$ Buy assumption, $F$ is independent of the choice of $\gamma$. Your job now is to show that $dF=w.$